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Unformatted text preview: L. Vandenberghe 10/26/10 EE103 Midterm solutions Problem 1 (10 points). A square matrix A is called normal if AA T = A T A . Show that if A is normal and nonsingular, then the matrix Q = A 1 A T is orthogonal. Solution. We show that Q T Q = I : Q T Q = ( A 1 A T ) T ( A 1 A T ) = AA T A 1 A T = A ( AA T ) 1 A T = A ( A T A ) 1 A T = AA 1 A T A T = I. On line 2 we use ( BC ) T = C T B T . On lines 3 and 5 we use the property that ( BC ) 1 = C 1 B 1 if B and C are invertible. On line 4 we use the definition of normal matrix. Problem 2 (10 points). Let a = ( a 1 , a 2 , a 3 ), b = ( b 1 , b 2 , b 3 ), c = ( c 1 , c 2 , c 3 ), d = ( d 1 , d 2 , d 3 ) be four given points in R 3 . The points do not lie in one plane. Algebraically, this can be expressed by saying that the vectors b a , c a , d a are linearly independent, i.e. , y 1 ( b a )+ y 2 ( c a )+ y 3 ( d a ) = 0 only if y 1 = y 2 = y 3 = 0. Suppose we are given the distances of a point x = ( x 1 , x 2 , x 3 ) to the four points: bardbl x a bardbl = r a , bardbl x b bardbl = r b , bardbl x c bardbl = r c , bardbl x d bardbl = r d . Write a set of linear equations Ax = f , with A nonsingular, from which the coordinates x 1 , x 2 , x 3 can be computed. Explain why the matrix A is nonsingular. Solution. Squaring the norms gives four nonlinear equations bardbl x bardbl 2 2 a T x + bardbl a bardbl 2 = r 2 a bardbl x bardbl 2 2 b T x + bardbl b bardbl 2 = r 2 b bardbl x bardbl 2 2 c T x + bardbl c bardbl 2 = r 2 c bardbl x bardbl 2 2 d T x + bardbl d bardbl 2 = r 2 d . Subtracting the first equation from the three others gives 2( a b ) T x = r 2 b r 2 a + bardbl a bardbl 2 bardbl b bardbl 2 2( a c ) T x = r 2 c r 2 a + bardbl a bardbl 2 bardbl c bardbl 2 2( a d...
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This note was uploaded on 02/03/2011 for the course EE 103 taught by Professor Vandenberghe,lieven during the Spring '08 term at UCLA.
 Spring '08
 VANDENBERGHE,LIEVEN

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